Algebra II Problems?1?

Question

1. 2x - y = 3z -3 
3x + 2y = z - 1 
x + 3y = z - 10 

2. 3u - 2v + w = 4 
5u + 3v - w = -2 
2u + ____ w =1 

3. (1/r) + (3/s) - (2/t) = 1 
(2/r) + (3/s) - (4/t) = 1 
(1/r) - (6/s) - (6/t) = 0 

Answer 1

1)

Write the system of equations in the form ax + by + cz = d.

2x - y - 3z = - 3 ---> (1)

3x + 2y - z = - 1 ---> (2)

x + 3y - z = - 10 ---> (3)

Use the elimination method to make a system of two equations in two variables.

Multiply each side of equation(3) by negative 2.

- 2x - 6y + 2z = 20 ---> (4)

Multiply each side of equation(3) by negative 3.

- 3x - 9y + 3z = 30 ---> (5)

 

Now the pair of  equations (1),(4) and (2),(5) contains opposite coefficient of x - variable.

To eliminate the x variable, add the equations (1) and (4).

(2x - y - 3z) + ( - 2x - 6y + 2z) = - 3 + 20

2x - y - 3z - 2x - 6y + 2z = 17

- 7y - z = 17 ---> (6)

To eliminate the x variable, add the equations (2) and (5).

(3x + 2y - z) + (- 3x - 9y + 3z) = - 1 + 30

3x + 2y - z - 3x - 9y + 3z = 29

- 7y + 2z = 29 ---> (7)

To eliminate the y variable subtract equations(6) from (7).

(- 7y + 2z) - ( - 7y - z) = 29 - 17

3z = 12

z = 4

Substitute the z value in equation(6).

- 7y - 4 = 17

7y = - 21

y = - 3

Substitute the y, z values in equation(3).

x + 3(- 3) - 4 = - 10

x - 9 - 4 = - 10

x = 3

Solutions are x = 3, y = - 3 and z = 4.

Answer 2

3)The system of equations are 1/r + 3/s - 2/t = 1 ,

2/r  + 3/s - 4/t = 1,

1/r - 6/s - 6/t = 0

Let 1/r = a, 1/s = b and 1/t = c.

Now the equations are a + 3b - 2c = 1 ---> (1)

2a + 3b - 4c = 1 ---> (2)

a - 6b - 6c = 0 ---> (3)

Use the elimination method to make a system of two equations in two variables.

Multiply each side of equation(3) by negative 2.

- 2a + 12b + 12c = 0 ---> (4)

Multiply each side of equation(1) by negative 2.

- 2a - 6b + 4c = - 2 ---> (5)

 

Now the pair of  equations (2),(4) and (2),(5) contains opposite coefficient of a - variable.

To eliminate the a variable, add the equations (2) and (4).

(2a + 3b - 4c) + (- 2a + 12b + 12c) = 1 - 0

2a + 3b - 4c - 2a + 12b + 12c = 1

15b + 8c = 1 ---> (6)

To eliminate the a variable, add the equations (2) and (5).

(2a + 3b - 4c) + (- 2a - 6b + 4c) = 1 + (- 2)

2a + 3b - 4c - 2a - 6b + 4c = 1 - 2

- 3b  = - 1

b = 1/3

Substitute the b value in equation (6).

15(1/3) + 8c = 1

5 - 1 = - 8c

c = - 4/8

c = -1/2

Substitute the b,c values in equation (1).

a + 3(1/3) - 2(-1/2) = 1

a + 1 + 1 = 1

a = - 1

Solutions are r = - 1, s = 3 and t = -2.